Question

if 9.2 g of calcium react completely with excess aluminum chloride how many grams of aluminum would be produced?
Ca + AlCl3 -> CaCl2 + Al

Answer 1

Explanation:

3Ca(s) + 2AlCl3(aq) -> 3CaCl2(aq) + 2Al(s)

According to the question, Ca is the limiting reactant.

Therefore, we equate Ca to Aluminium which is the product whose mass we want to find

Molar mass of Ca- 40g/mol

". ". of Al- 27g/mol

3Ca --> 2Al

3×40 --> 2×27

9.2 --> x

x = 9.2×2×27= 496.8÷120=4.14